The quadratic $x^2-3x+9=x+41$ has two solutions.  What is the positive difference between these solutions?
First we bring $x$ to the left side to get  \[x^2-4x+9=41.\]We notice that the left side is almost the square $(x-2)^2=x^2-4x+4$. Subtracting 5 from both sides lets us complete the square on the left-hand side, \[x^2-4x+4=36,\]so  \[(x-2)^2=6^2.\]Therefore $x=2\pm6$.  The positive difference between these solutions is $8-(-4)=\boxed{12}$.